How to split a quartic into two quadratics? I have a quartic in $\Bbb Z[x]$ with very large coefficients that I know splits into two quadratics in $\Bbb Z[x]$. What is the best way to do find the quadratics?
 A: Use the coefficients relation, Look here
For example, you want to split: 
$8+10x+9x^2+3a^3+x^4 = (x^2+px+q)(x^2+rx+s)$
The polynomial coefficients must be equal, so:
 (1) $3 = p+q$
 (2) $9 = q+s+pr$
 (3) $10 = ps+qr$
 (4) $8 = qs$
 this is a system of equtions on $\mathbb{Z}$...
 multipliying (1) by $s$ and by $r$
 (5) $3s = ps+qs=ps+8$
 (6) $3r = pr+qr$
 (3)-(5) :
 $10-3s = ps+qr-ps-qs=qr-8$
 ($6'$) $18 = 3s + qr$
 (2)-(6) :
 $9-3r = q+s-qr$
 (7) $qr=q+s-3r+9$
 substitute (7) in ($6'$) :
 $18=3s+q+s-3r+9$
 (8) $9=4s+q-3r$
 (1)-(2) :
 (9) $9=q+s+(3-q)r=q+s+3r-qr$
 (9)+(3):
 $19=q+s+3r-qr+ps+qr=q+s+3r+ps$
 and substituting $ps$ from (5):
 $19=q+s+3r+3s-8$
 (10) $27=q+4s+3r$
 (10)-(8):
 $18=6r$
 $r=3$
rewriting and substituing $r=3$ in (6),(8)
 (6) $18 = 3s + 3q$
 (8) $18=4s+q$
 (6)-3*(8): $18-3\cdot18=3s-12s$
 $-36=-9s$
 $s=4$
 (6) $18 = 3\cdot4 + 3q$
 $6 = 3q$
 $q=2$
 (1) $3 = p+q=p+2$
 $p=1$
 You can repeat this process for  any numbers, just make sure you don't build second order equation, otherwise you will get more than one solution which will be not integer
A: Since there are three ways of partitioning a four element set into two pairs, finding an integer factorisation up to the order of the factors is about as hard as finding an integer root to a cubic (an abelian cubic, in fact). Then choosing the order of the factors is as hard as solving a general quadratic, so computing a square root.
Let $L = \Bbb Q(A,B,C,D)$ and $K = Q^{S_4} = \Bbb Q (E_1,E_2,E_3,E_4)$
where $E_i$ are the $i$th symmetric polynomials in $A,B,C,D$.
Getting a factorisation means that you have integer values for $A+B,C+D,AB,CD$.
Let $H_1$ be the subgroup of $S_4$ fixing $A+B$, and $H_2$ be the subgroup fixing $U = AB+CD$. 
$H_1$ has order $4$ and is a subgroup of $H_2$ (order $8$), so the fundamental theorem of Galois theory says that
$[K : L^{H_2} = K(U)] = 3$ and $[L^{H_2} : L^{H_1} = K(A+B) = \Bbb Q(A+B,AB,C+D,CD)] = 2$.
More precisely, if we focus on polynomials in the roots with integer coefficients,
it turns out that $\Bbb Z[A,B,C,D]^{H_2} = \Bbb Z[E_i,U]$,
where $U$ satisfies the equation
$(U-(AB+CD))(U-(AC+BD))(U-(AD+BC)) = 0$.
After developing, this is 
$U^3 - E_2 U^2 + (E_1E_3-4E_4) U - (E_1^2E_4+E_3^2-4E_2E_4) = 0$,
so first you want to find the integer roots of this cubic.
Once you have an integer value for $U$, you have determined that $A$ goes with $B$ and $C$ goes with $D$, but you still haven't distinguished the two pairs from each other.
Then, $\Bbb Z[A,B,C,D]^{H_1} = \Bbb Z[E_i,A+B,AB] = \Bbb Z[A+B,AB,C+D,CD]$, and we have the following equations :  
$((A+B)-(C+D))^2 = 4U + E_1^2 - 4E_2 \\
((A+B)-(C+D))(AB-CD)  = E_1U - 2E_3 \\
(AB-CD)^2 = U^2 - 4E_4$    
so you can find the integer values of $A+B-C-D$ and $AB-CD$ by computing an integer square root, and then doing an integer division with the second equation. Note that if $A+B=C+D$ then you need the third equation to find $AB-CD$, and if $AB=CD$ you need the first equation to find $A+B-C-D$, so to cover all the cases, you need all three of them.
And finally, you get $A+B,C+D,AB,CD$ after computing $\frac 12 (E_1 \pm (A+B-C-D))$ and $\frac 12 (U \pm (AB-CD))$.
Once you have those values, you have the factorisation $(X^2-(A+B)X+AB)(X^2-(C+D)X+CD)$ that you wanted.
A: Can we assume that all resulting coefficients are integers?  If so, use the rational root theorem to identify a list of possible roots and test them using the factor theorem.  Once you found one root, you can divide it from the quartic to get a cubic.  Repeat one more time and you have your quadratic.  It's not the most efficient if your coefficient are really big.
